Question

The value of $\frac{{}^n{C}_0}{1\cdot 2}-\frac{{}^n{C}_1}{2\cdot 3}+\frac{{}^n{C}_2}{3\cdot 4}+\cdots +(-1{)}^n\frac{{}^n{C}_n}{(n+1)\cdot (n+2)}$ is

• A. $\frac{1}{n+2}$
• B. $\frac{1}{n+1}$
• C. $\frac{1}{n+3}$
• D. $\frac{1}{(n+2)(n+1)}$

Answer 1

The correct option is A $\frac{1}{n+2}$

Let
$S=\frac{{}^n{C}_0}{1\cdot 2}-\frac{{}^n{C}_1}{2\cdot 3}+\frac{{}^n{C}_2}{3\cdot 4}+\cdots +(-1{)}^n\frac{{}^n{C}_n}{(n+1)\cdot (n+2)}$
Considering,
$(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+\cdots +{}^n{C}_n{x}^n$
Integrating w.rt. $x$,
$\frac{(1+x{)}^{n+1}}{n+1}={}^n{C}_{0}x+{}^n{C}_1\frac{x^2}{2}+\cdots +{}^n{C}_n\frac{x^{n+1}}{n+1}+k$
Putting
$x=0\Rightarrow k=\frac{1}{n+1}$
$\frac{(1+x{)}^{n+1}}{n+1}={}^n{C}_{0}x+{}^n{C}_1\frac{x^2}{2}+\cdots +{}^n{C}_n\frac{x^{n+1}}{n+1}+\frac{1}{n+1}$
Again integrating w.r.t. $x$,
$\frac{(1+x{)}^{n+2}}{(n+1)(n+2)}={}^n{C}_0\frac{x^2}{1\cdot 2}+{}^n{C}_1\frac{x^3}{2\cdot 3}+\cdots +{}^n{C}_n\frac{x^{n+2}}{(n+1)(n+2)}+\frac{x}{n+1}+k_1$
Putting
$x=0\Rightarrow k_1=\frac{1}{(n+1)(n+2)}$
$\frac{(1+x{)}^{n+2}}{(n+1)(n+2)}={}^n{C}_0\frac{x^2}{1\cdot 2}+{}^n{C}_1\frac{x^3}{2\cdot 3}+\cdots +{}^n{C}_n\frac{x^{n+2}}{(n+1)(n+2)}+\frac{x}{n+1}+\frac{1}{(n+1)(n+2)}$
Now, putting $x=-1$,
$0=S-\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}\Rightarrow S=\frac{1}{n+2}$